Latus Rectum of Locus of Centroid of Inscribed Triangle in Parabola

To find three times the length of the latus rectum of the conic $C_0$, we first determine the equation of the locus of the centroid of $\Delta ABC$.

The given parabola is: $y^2 = 16x$

Comparing with the standard equation $y^2 = 4ax$, we have $a = 4$. Any general point on the parabola can be represented in parametric form as: $(4t^2, 8t)$

Given that the vertex $B$ containing the right angle is $(4, 8)$, its parameter $t_B$ is obtained by setting $8t = 8 \implies t_B = 1$.

Let the parameters for the other two vertices $A$ and $C$ be $t_1$ and $t_2$ respectively, where $t_1, t_2 \neq 1$. Thus, $A = (4t_1^2, 8t_1)$ and $C = (4t_2^2, 8t_2)$.

The slope of line $AB$ is given by: $m_{AB} = \frac{8t_1 - 8}{4t_1^2 - 4} = \frac{8(t_1 - 1)}{4(t_1 - 1)(t_1 + 1)} = \frac{2}{t_1 + 1}$

Similarly, the slope of line $BC$ is: $m_{BC} = \frac{2}{t_2 + 1}$

Since $\Delta ABC$ is a right-angled triangle with the right angle at $B$, the lines $AB$ and $BC$ are perpendicular ($AB \perp BC$). Therefore: $m_{AB} \cdot m_{BC} = -1$ $\left(\frac{2}{t_1 + 1}\right) \cdot \left(\frac{2}{t_2 + 1}\right) = -1$ $4 = -(t_1 + 1)(t_2 + 1)$ $4 = -(t_1 t_2 + t_1 + t_2 + 1)$ $t_1 t_2 + t_1 + t_2 + 5 = 0 \quad \text{--- (1)}$

Let $(h, k)$ be the coordinates of the centroid of $\Delta ABC$: $h = \frac{4 + 4t_1^2 + 4t_2^2}{3} \implies t_1^2 + t_2^2 = \frac{3h - 4}{4} \quad \text{--- (2)}$ $k = \frac{8 + 8t_1 + 8t_2}{3} \implies t_1 + t_2 = \frac{3k - 8}{8} \quad \text{--- (3)}$

From equation (1), we can express $t_1 t_2$ in terms of $k$: $t_1 t_2 = -5 - (t_1 + t_2) = -5 - \frac{3k - 8}{8} = \frac{-32 - 3k}{8} \quad \text{--- (4)}$

Using the algebraic identity $t_1^2 + t_2^2 = (t_1 + t_2)^2 - 2t_1 t_2$, we substitute equations (2), (3), and (4): $\frac{3h - 4}{4} = \left(\frac{3k - 8}{8}\right)^2 - 2\left(\frac{-32 - 3k}{8}\right)$ $\frac{3h - 4}{4} = \frac{(3k - 8)^2}{64} + \frac{32 + 3k}{4}$

Multiplying the entire equation by $64$: $16(3h - 4) = (3k - 8)^2 + 16(32 + 3k)$ $48h - 64 = (9k^2 - 48k + 64) + (512 + 48k)$ $48h - 64 = 9k^2 + 576$ $9k^2 = 48h - 640$ $k^2 = \frac{48}{9}h - \frac{640}{9}$ $k^2 = \frac{16}{3}\left(h - \frac{40}{3}\right)$

Replacing $(h, k)$ with $(x, y)$, the locus $C_0$ is the parabola: $y^2 = \frac{16}{3}\left(x - \frac{40}{3}\right)$

This is a parabola of the form $Y^2 = 4aX$, where the length of the latus rectum $L$ is the coefficient of the linear term in $x$: $L = \frac{16}{3}$

We are required to find three times the length of the latus rectum of $C_0$: $3 \times L = 3 \times \frac{16}{3} = 16$